May 12th, 2009, 4:56 pm
Ok, we'll go to two factors. I copied and pasted my earlier post and made the necessary modifications:-----------------------------------------------------------------------------------------------------------------------------------Well, for general 2-factor, we are talking about a PDE(A) f_t = a_ij(t,x) f_ij + b_i(t,x) f_i - c(t,x) fwith f(0,x) a given (non-negative) function.My notation: x = (x1,x2) is a two-vector. f_i = Partial f/Partial xiThere is an implied summation on repated indices.If I write b( ) with no subscript, I mean the 2-vector (b1,b2).If write a( ) with no subscripts, I mean the 2 x 2 matrix [a_ij] Let x=B denote any point on a degenerate boundary -- such points are defined by det[a(t,B)] = 0 there. Again B is a 2-vector. I have in mind x = (S,V) for stochastic vol. models, so state space is the positive real first quadrant: x > 0For Example Cases 1,2,3 (see bottom of the post), there are two degenerate boundariesBoundary 1: x1 = 0 (S=0) and Boundary 2: x2 = 0 (V=0 or S2=0 for Example Case 3)So, x=B means x is a point on Boundary 1 or x is a point on Boundary 2. Assume the other boundaries are well-understood and not an issue.So, the question becomes, under this premise, when does it make sense to simply employ(*) f_t = a_ij(t,B) f_ij + b_i(t,B) f_i - c(t,B) fon a degenerate boundary, where certain components of [a_ij(t,B)] may vanish (or none vanish, but the det. does) I want the killing term c(t,x) to not be singular, so,for simplicity, let's assume (i) 0 <= c(t,x) < infinity for all (t,x).Take Boundary 1 and assume the associated SDE for the first component of X, at the boundary, reduces todX_1 = b_1(t,0,X2) dt Similarly, take Boundary 2 and assume the associated SDE for the second component of X, at the boundary, reduces todX_2 = b_2(t,X1,0) dt This is not the most general thing that can happen when det(a) = 0, but it is a good start. Then, the process can "keep running" when Boundary i is hit if and only if b_i > 0 there, where i = 1,2Why? Because this means that once the boundaryis hit, the particle is simply returned to the interior of the state space x > 0. The probability ofbeing killed there is not an issue because c(t,B) is bounded. In addition, even if theboundary is never hit (entrance boundary, say), the process can still be started there.So, from this point of view, everything proceeds nicely.This suggests to me that ======================================================A sufficient set of conditions for a well-posed numerical PDE problem (A), using the BC (*) at a degenerate boundary x=B is that:(i) det[a(t,B)] = 0, and(ii) n . b(t,B) > 0, and(iii) 0 < = c(t,B) < infinityHere n is the inward-pointing normal to the boundary at x=B and "." is the dot product.In other words, the drift will return the particle to the interior of the state space(Of course, there are additional standard regularity conditions, like a(t,x) > 0 for x > 0, etc.)==================================================================Example Cases: 1. Heston: Boundary1. (S=0): b_1(t,S=0,V) = 0, so the sufficiency condition (ii) is violated.Boundary2. (V=0): b_2(t,S,V=0) > 0, so the sufficiency condition (ii) is OK, and (*) may be applied there. 2. SABR: Boundary1. (S=0): b_1(t,S=0,V) = 0, so the sufficiency condition (ii) is violated. Boundary2. (V=0): b_2(t,S,V=0) = 0, so the sufficiency condition (ii) is violated. 3. Multi-asset (n = 2) PDE with correlation Boundary1. (S1=0): b_1(t,S1=0,S2) = 0, so the sufficiency condition (ii) is violated. Boundary2. (S2=0): b_2(t,S1,S2=0) = 0, so the sufficiency condition (ii) is violated. 4. Convertible bonds -- what is the PDE?
Last edited by
Alan on May 11th, 2009, 10:00 pm, edited 1 time in total.